Question

${\sin }^6\frac{\pi }{49}+{\cos }^6\frac{\pi }{49}-1+3{\sin }^2\frac{\pi }{49}{\cos }^2\frac{\pi }{49}$

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

Take $a={\sin }^2\frac{\pi }{49}$ and $b={\cos }^2\frac{\pi }{49}$

The given expression can be written as
$a^3+b^3-1+3ab$
$=(a+b)(a^2-ab+b^2)-1+3ab$
but $a+b=1$

So, the expression reduces to

$a^2-ab+b^2-1+3ab$

$=a^2+2ab+b^2-1$

$=(a+b{)}^2-1$

$=1-1=0$